Python | pycurl sends wildcard as path

Im currently writing a program which connects to a cam using RTSP. The python script is using pycurl.

I managed to make OPTIONS and DESCRIBE work. However SETUP is not working.

> Choose request or command: SETUP * Found bundle for host 0x7f8e70a0b650 [can pipeline] * Re-using existing connection! (#0) with host * Connected to ( port 554 (#0) > SETUP * RTSP/1.0 CSeq: 3 Transport: RTP/AVP;unicast;client_port=38274-38275 User-Agent: LibVLC/3.0.8 (LIVE555 Streaming Media v2018.02.18) Accept: */*  < RTSP/1.0 400 Bad Request < Allow: OPTIONS, DESCRIBE, SETUP, TEARDOWN, PLAY, PAUSE, GET_PARAMETER, SET_PARAMETER,USER_CMD_SET <  * The CSeq of this request 3 did not match the response 0 * Connection #0 to host left intact Traceback (most recent call last):   File "/Volumes/DATA/htdocs/forensik/forensik_curl/venv/", line 209, in <module>     rtsp.handle(cmd)   File "/Volumes/DATA/htdocs/forensik/forensik_curl/venv/", line 99, in handle     self.sendRequest(pycurl.RTSPREQ_SETUP, header)   File "/Volumes/DATA/htdocs/forensik/forensik_curl/venv/", line 53, in sendRequest     self.crl.perform() pycurl.error: (85, 'The CSeq of this request 3 did not match the response 0') * Closing connection 0 

I think the problem is that there is a wildcard * in the request path


To set the URL is use

pycrl.setopt(pycrl.URL, rtsp://

Any idea what I can do ?

Thanks in advance!

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